a.
To calculate:The equation of line that is tangent to the given curve at the given point.
a.
Answer to Problem 26E
The equation of tangent is
Explanation of Solution
Given information:The equation of the curve is
Formula used:
Also product, quotient and chain rule of derivatives are used.
Calculation:
The curve equation is
And the point at which the tangent equation is to be determined is
By differentiating equation (1) ,
Here
The slope point form of the tangent line is
Hence the equation of tangent is
b.
To calculate:The equation of line that is normal to the given curve at the given point.
b.
Answer to Problem 26E
The equation of normal to the curve is
Explanation of Solution
Given information:The equation of the curve is
Formula used:
The slope of normal to the curve at a given point is
Where m is the slope of the tangent to the curve at that point.
Calculation:
Here
Then the equation of normal at
Hence the equation of normal to the curve is
Chapter 4 Solutions
Calculus: Graphical, Numerical, Algebraic: Solutions Manual
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